My third "alternate ending" idea in 3 days (This one's for Roux)

[blah]

I don't know if this is a new idea at all, but I haven't seen it on Gilles' website, so I'm gonna post it here. It's an extremely trivial method, really.

Rather than always solving the UR and UL edges first (because that's what you were taught to do), solve any two opposite edges first, it's called freestyle

Case 1: 2 edges that are supposed to be opposite each other are already opposite each other, i.e. either UR opposite UL, or UF-UB, or DF-DB. Solution: (AUF), M2, AUF, M2, solved.

Case 2: If Case 1 doesn't occur, Case 2 definitely will (sometimes Case 2 even occurs simultaneously with Case 1 since they're not mutually exclusive). Of any 2 edges that are supposed to be opposite each other, one's on D while the other's on U. Solution: (AUF), M' U2 M/M', AUF, M2, solved.

Very easy, very intuitive, very fast. Very easy recognition for Case 2, because the D stickers will stand out among all the U stickers. I just do Case 2 if I don't spot Case 1 in time, only takes an extra U2, which is definitely faster to execute than to try locating 2 opposite edges pieces out of the 6.

After this, you'll get an EPLL case 11/36 (30.6%) of the time, or a 4-move M-slice edge permutation 11/18 (61.1%) of the time, or a skip 1/12 (8.3%) of the time.

Recognition really shouldn't be a problem if you've been Roux-ing long enough, I started messing with Roux today and I can recognize cases instantly What I do: Do an M move to solve centers, then do the usual algs. There's also M' Uw2 M' Uw2 (and all its inverses and mirrors and their inverses) in case you never knew about it

Edit:
New idea to deal with those annoying EPLL cases that are so long compared to those M-slice edge permutations: Choose only to solve either UR-UL or UF-UB after orienting edges. They should fall under Cases 1 and/or 2 every single time, except for these cases:
1. DF-DB swap, 2 adjacent U edgess swap
2. DF-DB swap, U edges 4-cycle (6 cases, I think)

Solution: Learn algs and get a 1-look 6E4C

 

[AvGalen]

If it wasn't for the quality of your ideas I would have made a joke about the 3 ideas.

(blah blah blah)

 

[blah]

If it wasn't for the quality of your ideas I would have made a joke about the 3 ideas.

(blah blah blah)

You could still make that joke even with the quality of my ideas

 

[Athefre]

I'm sure most "Roux" users have considered this, including myself. I've thought about it a lot and it doesn't seem worth it to me. Notes:

a. Recognition for Step 4c when doing what you call "freestyle" takes a slight amount more thinking time than a normal Step 4c.
b. The move count is the same as doing a normal solve and it isn't any faster. You have to hope that you get better (faster) at recognizing 2 opposite edges and knowing exactly how to react.

My best average of 12 for the LSE is 3.01 seconds. I don't see how any other technique could lower that, or any reason why someone should practice something that takes more time to recognize.

 

[blah]

b. The move count is the same as doing a normal solve and it isn't any faster.

It is lower on average, by barely anything though You're right, it isn't any faster.

Probability of UR and UL pieces being adjacent on U: 4/15 (6.75 moves including AUF)
Probability of UR and UL pieces being opposite on U: 2/15 (1.5 moves including AUF)
Probability of UR and UL being on U and D: 8/15 (5.5 moves including AUF)
Probability of UR and UL being on D: 1/15 (1.75 moves including AUF)
Overall average number of moves: 5.05 STM

Edit: I redid everything, had major miscalculations just now.

 

[Athefre]

b. The move count is the same as doing a normal solve and it isn't any faster.

It is lower on average, by barely anything though You're right, it isn't any faster.

Probability of UR and UL pieces being adjacent on U: 4/15 (6.75 moves including AUF)
Probability of UR and UL pieces being opposite on U: 2/15 (1.5 moves including AUF)
Probability of UR and UL being on U and D: 8/15 (5.5 moves including AUF)
Probability of UR and UL being on D: 1/15 (1.75 moves including AUF)
Overall average number of moves: 5.05 STM

Edit: I redid everything, had major miscalculations just now.

What would you say is the total move count average for "freestyle"? I average 13 moves for normal complete LSE (Step 4a, 4b, and 4c). Would you say "freestyle" is less than that?

 

[cubacca1972]

I like the stuff in Roux's method, but get a little tripped up with the orientation of the remaining edge cubies stuff. This is mostly to do with how to handle the cases where one of the UR or UL (or both) edges are in the UR or UL position when you are about to do one of the algs. They don't behave the same way as the M edges. If an M edge is flipped in the UR position, the move M or M' changes it to oriented. Conversely, if The UR edge is in UR or UL position and is flipped, the move M or M' doesn't change its orientation. This makes it difficult for me to predict what will happen next, and slows down my recognition time.

I prefer to solve corners first, but was intrigued by the orientation step, so I played with it for a bit. I looked at Jelinek's 2.5 step solution http://rubikscube.info/lastsix2look.php
And liked that idea too, except for the number of algorithms, as this clashes with my natural laziness.

Instead, I made up my own system (more likely rediscovered an old idea), and found my own algorithms for it.

First, I place the UL and UR cubies in the DF and DB position, disregarding their orientation or which one is placed in either spot (maximum 3 moves). Next, I turn U, so that the cube matches one of eleven orientation pattens (3 out of the 11 have all 4 edge cubies in the U layer all flipped or all oriented, so no U turn needed to set up).

Once the algorithm is executed, all edge cubies are oriented, the UL and UR cubies are oriented and still in the D layer, and can be solved with U move, M2, and a U move. The nice thing is that in one out of 12 cases, the M layer gets solved when you solve UL and UR.

The average number of moves for the 11 algs is 9.18 moves. Five of the algs are 7 moves, one alg is 9 moves, four of the algs are 11 moves, and the worst case is 13 moves. Each alg ends with an M move, and should lead to cases (1/4 of the time) where UR and UL cubies can be direct solved rather than getting parked in the D layer. This would probably reduce the total moves needed a fair bit.

 

[Athefre]

I like the stuff in Roux's method, but get a little tripped up with the orientation of the remaining edge cubies stuff. This is mostly to do with how to handle the cases where one of the UR or UL (or both) edges are in the UR or UL position when you are about to do one of the algs. They don't behave the same way as the M edges. If an M edge is flipped in the UR position, the move M or M' changes it to oriented. Conversely, if The UR edge is in UR or UL position and is flipped, the move M or M' doesn't change its orientation. This makes it difficult for me to predict what will happen next, and slows down my recognition time.

Recognition is the easiest part of the LSE. All you do is make sure all edges have the color of the U corners facing the U or D layer, you don't have to locate the UL/UR edges unless you are doing non-matching blocks.

I prefer to solve corners first, but was intrigued by the orientation step, so I played with it for a bit. I looked at Jelinek's 2.5 step solution http://rubikscube.info/lastsix2look.php
And liked that idea too, except for the number of algorithms, as this clashes with my natural laziness.

Instead, I made up my own system (more likely rediscovered an old idea), and found my own algorithms for it.

First, I place the UL and UR cubies in the DF and DB position, disregarding their orientation or which one is placed in either spot (maximum 3 moves). Next, I turn U, so that the cube matches one of eleven orientation pattens (3 out of the 11 have all 4 edge cubies in the U layer all flipped or all oriented, so no U turn needed to set up).

Once the algorithm is executed, all edge cubies are oriented, the UL and UR cubies are oriented and still in the D layer, and can be solved with U move, M2, and a U move. The nice thing is that in one out of 12 cases, the M layer gets solved when you solve UL and UR.

The average number of moves for the 11 algs is 9.18 moves. Five of the algs are 7 moves, one alg is 9 moves, four of the algs are 11 moves, and the worst case is 13 moves. Each alg ends with an M move, and should lead to cases (1/4 of the time) where UR and UL cubies can be direct solved rather than getting parked in the D layer. This would probably reduce the total moves needed a fair bit.

This is interesting, it would probably go in the same category as Kenneth's E15. Would you be willing to do an average of at least 12 solves and give us the move count?

UL/UR placement: 3
Edge orient: 9
Edge permute: At least 4, permuting the final M edges has an average of 4 moves by itself and I'm not sure what techniques you use to place the UL/UR edges.

And of course there are some move cancellations in-between each substep.

I like the idea, the only problem I see is that you have to locate the UL/UR edges. This creates a pause after completing the corners, compared to a normal solve where you finish the corners and the solve keeps flowing because edge orientation recognition is instant. Hopefully with practice you could locate both edges while performing the corner sequence.

When you place the UL/UR edges in this idea, does M center orientation matter? I'm guessing it doesn't because if a UL/UR edge is at DB and the other at UB, it takes 4 moves to bring it to DF if centers matter. If they don't, it's 1 move.

 

[cubacca1972]

This is interesting, it would probably go in the same category as Kenneth's E15. Would you be willing to do an average of at least 12 solves and give us the move count?

I did an average of 10. I use ortega to solve corners, then place edges one at a time. This will lead to a higher move count I suppose, so I separated the moves as follows:

first number represents all corners and L and R edges solved, plus the turn necessary to align the R layer with the L layer, so that there are two unsolved edge spaces at LU and RU.

Second number represents all the moves needed to set up the pattern, plus one of the eleven algorithms.

Third number represents solving LU and RU, solving M layer, and aligning L, M, and R layers (I don't use Roux, so my RU edge cubie shares a facelet color with my LU cubie only 1/4 of the time).

1. 44+8+8=58
2. 51+15+5=71 (I think I miscounted, but 15+5 is accurate)
3. 37+10+8=55
4. 37+10+6=53
5. 42+10+4=56 (M was solved when I finished orientation + solve LU, RU)
6. 41+13+7=61
7. 43+9+7=59
8. 35+6+4=45 (Edges were oriented once I slotted UL and UR in D layer)
9. 42+15+4=61
10. 45+8+6=59

UL/UR placement: 3
Edge orient: 9
Edge permute: At least 4, permuting the final M edges has an average of 4 moves by itself and I'm not sure what techniques you use to place the UL/UR edges.

When my orientation step is done, the UR and UL cubies are in the D layer. IF the U layer happens to be in the solved position, UR and UL would be solved with either U M2 U', or U' M2 U, leaving the M layer oriented but not solved. There would be move cancellation potential at the very end of each orientation algorithm 1/4 of the time, dependent of course on the position of the U layer during the last M move in the algorithm.

I like the idea, the only problem I see is that you have to locate the UL/UR edges. This creates a pause after completing the corners, compared to a normal solve where you finish the corners and the solve keeps flowing because edge orientation recognition is instant. Hopefully with practice you could locate both edges while performing the corner sequence.

Locating isn't too bad, as there are only two cubies to hunt. You could certainly track at least one of them while doing the last moves pre-orientation step with either Roux or corners first.

When you place the UL/UR edges in this idea, does M center orientation matter? I'm guessing it doesn't because if a UL/UR edge is at DB and the other at UB, it takes 4 moves to bring it to DF if centers matter. If they don't, it's 1 move.

The M center orientation doesn't matter, so it is only 3 moves maximum.

 

[cubacca1972]

As this is a Roux thread, I did 12 solves using corners first, but taking care to ensure that my UL and UR cubies belong to the same U face (eg, both share yellow facelets. Here is the total move counts, where I tried to incorporate move cancellations whenever I saw them. Note that these move counts only represent the number of moves needed to set up the LU and RU edge cubies in the D layer, turn U to get the cube to match one of 11 cases, and to completely solve the cube.

20 17 19 19 20 17 (15) (12) 20 15 16 19 Avg= 18.2

I am aware that my total solve move count is high, but I am not likely to change my method leading up to the orientation stage, as I am lazy. I just thought I would throw my orientation method into the mix as a point of potential interest.

 

[Athefre]

As this is a Roux thread, I did 12 solves using corners first, but taking care to ensure that my UL and UR cubies belong to the same U face (eg, both share yellow facelets. Here is the total move counts, where I tried to incorporate move cancellations whenever I saw them. Note that these move counts only represent the number of moves needed to set up the LU and RU edge cubies in the D layer, turn U to get the cube to match one of 11 cases, and to completely solve the cube.

20 17 19 19 20 17 (15) (12) 20 15 16 19 Avg= 18.2

I am aware that my total solve move count is high, but I am not likely to change my method leading up to the orientation stage, as I am lazy. I just thought I would throw my orientation method into the mix as a point of potential interest.

That's about 4-5 moves more than what Gilles' method averages but it might could be shortened with practice. Do you mind posting your 11 orientation solutions that keep UL/UR at DF/DB so I (and I guess others) can see what I average? When I was solving your way sometimes it felt like I was wasting moves. I had one case where I had 3 bad edges on U and the other at DF (a UL/UR edge!) and the only solution I saw (if I wanted to keep UL/UR where they are) was a variation of M'UMU'M'U2M. Instead I kept wanting to switch back to the way I'm used to solving so I could just orient then solve UL/UR.

 

[cubacca1972]

That's about 4-5 moves more than what Gilles' method averages but it might could be shortened with practice. Do you mind posting your 11 orientation solutions that keep UL/UR at DF/DB so I (and I guess others) can see what I average?

M U' M U M' U2 M U' M U' M (11,12)
M U M' U2 M U M' (7,8)
M' U2 M' U2 M' U' M U2 M U2 M (11,15)
M' U M' U2 M' U M' U M U' M' (11,12)
M' U M U' M' U2 M (7,8)
M U' M U2 M U' M U' M' U M (11,12)
M U' M' U M U2 M' (7,8)
M' U' M' U M U' M (7,7)
M' U2 M' U2 M U' M' U' M2 (9,12)
M U M U M U M (7,7)
M' U M' U M' U M2 U' M U2 M' U' M (13,15)

Sorry that I don't have diagrams. I use a printout of these algs with hand drawn diagrams. Just take a solved cube, do U M2 to get UL and UR in the D layer with all 6 edge cubies oriented, then do the inverse of each algorithm to derive the starting pattern.

When I was solving your way sometimes it felt like I was wasting moves. I had one case where I had 3 bad edges on U and the other at DF (a UL/UR edge!) and the only solution I saw (if I wanted to keep UL/UR where they are) was a variation of M'UMU'M'U2M. Instead I kept wanting to switch back to the way I'm used to solving so I could just orient then solve UL/UR.

The way I see it, the more tools you have in your toolbox, the better. If you happen to see an easy case for another system (LL one step corners, then one step edges-easy algs such as 3 cycle edges) then I would go for it.

There is a very real possibility that I am not fully grasping the Roux method of orientation.

Does M center orientation matter?

Does it matter where the UL and/or UR edges are when you do one of the iterative algorithms?

The answers to the above questions might help me understand the orientation system better.

When I looked at the patterns on the site, I figured that you could derive full algorithms (orient all 6 edge cubies) by staring with one pattern, setting up the next with a U move, doing the next move, etc.

Much to my dismay, any given pattern does not transform to the identical new pattern in every case. Therefore I couldn't just direct solve from one starting pattern. It ended up being look-adjust U-move-look-adjust U-move-etc.

If this is the way the system is supposed to go, then it is likely that it just doesn't mesh well with my idiosyncrasies.

With Roux, once the edge cubies are oriented, there are 30 different positions where UL and UR can be, one of which is in the solved position. With my system, there are 2 positions where UL and UR will be, both of which can be solved with 3 moves or less, depending on move cancellation potential.

Please correct me if I am misinterpreting the Roux method, as I am talking about my understanding of it.

 

[Athefre]

M U' M U M' U2 M U' M U' M (11,12)
M U M' U2 M U M' (7,8)
M' U2 M' U2 M' U' M U2 M U2 M (11,15)
M' U M' U2 M' U M' U M U' M' (11,12)
M' U M U' M' U2 M (7,8)
M U' M U2 M U' M U' M' U M (11,12)
M U' M' U M U2 M' (7,8)
M' U' M' U M U' M (7,7)
M' U2 M' U2 M U' M' U' M2 (9,12)
M U M U M U M (7,7)
M' U M' U M' U M2 U' M U2 M' U' M (13,15)

Sorry that I don't have diagrams. I use a printout of these algs with hand drawn diagrams. Just take a solved cube, do U M2 to get UL and UR in the D layer with all 6 edge cubies oriented, then do the inverse of each algorithm to derive the starting pattern.

Thanks, I'll study them, make an average, then post my results.

The way I see it, the more tools you have in your toolbox, the better. If you happen to see an easy case for another system (LL one step corners, then one step edges-easy algs such as 3 cycle edges) then I would go for it.

That's part of the reason I asked for your solutions, I'm going to see how useful it could be to partially add them to my normal solves (you get the credit if I use them in my solves of course . I like the idea.

There is a very real possibility that I am not fully grasping the Roux method of orientation.

Does M center orientation matter?

If the L/R blocks match, orientation matters if you want to be able to recognize the edge orientation quickly. If the centers aren't oriented you have to actually find UL/UR to see their orientation. Do this to a solved cube (orange/red on top and white/yellow on the sides so my description works):

M'U2M2UMU2MUM2U'M'UMU2

Notice that the centers aren't oriented, so you have to make sure all edges on U have green/blue facing the U-layer and both edges on D have to have green/blue facing the D-layer.

Now that you know that, you can see automatically that the edges a UL and UR are misoriented and the edge at UF is oriented correctly. But what about UB, DF, and DB? You have to actually tilt the cube and look at each edge because the edges that are supposed to be at UL/UR (the white/yellow ones) do not have green/blue on them.

So, DB is oriented correctly, UB is misoriented, and DF is misoriented. So to orient it's a simple U2M'U'M. So do U2M'UM backwards on a cube with centers oriented and you'll see that it's much easier to recognize (no cube tilting).

Does it matter where the UL and/or UR edges are when you do one of the iterative algorithms?

Nope.

The answers to the above questions might help me understand the orientation system better.

When I looked at the patterns on the site, I figured that you could derive full algorithms (orient all 6 edge cubies) by staring with one pattern, setting up the next with a U move, doing the next move, etc.

Much to my dismay, any given pattern does not transform to the identical new pattern in every case. Therefore I couldn't just direct solve from one starting pattern. It ended up being look-adjust U-move-look-adjust U-move-etc.

If this is the way the system is supposed to go, then it is likely that it just doesn't mesh well with my idiosyncrasies.

With Roux, once the edge cubies are oriented, there are 30 different positions where UL and UR can be, one of which is in the solved position. With my system, there are 2 positions where UL and UR will be, both of which can be solved with 3 moves or less, depending on move cancellation potential.

Please correct me if I am misinterpreting the Roux method, as I am talking about my understanding of it.

Yeah, that's how it works, at first it seems stupid and wasteful. But once you use it a while you'll naturally have all of the U adjustments memorized and you'll realize that the moves are exactly same as memorized sequences but much more free. Having them separated like that allows you to control where UL/UR goes while you are orienting. Why have M'UM'U2M'UM' memorized when you can do a variation of M'UM' U2 M'U'M so that you can place UL/UR while orienting.

 

[cubacca1972]

If the L/R blocks match, orientation matters if you want to be able to recognize the edge orientation quickly. If the centers aren't oriented you have to actually find UL/UR to see their orientation. Do this to a solved cube (orange/red on top and white/yellow on the sides so my description works):

M'U2M2UMU2MUM2U'M'UMU2

Notice that the centers aren't oriented, so you have to make sure all edges on U have green/blue facing the U-layer and both edges on D have to have green/blue facing the D-layer.

Now that you know that, you can see automatically that the edges a UL and UR are misoriented and the edge at UF is oriented correctly. But what about UB, DF, and DB? You have to actually tilt the cube and look at each edge because the edges that are supposed to be at UL/UR (the white/yellow ones) do not have green/blue on them.

So, DB is oriented correctly, UB is misoriented, and DF is misoriented. So to orient it's a simple U2M'U'M. So do U2M'UM backwards on a cube with centers oriented and you'll see that it's much easier to recognize (no cube tilting).

I think that if faced with a situation where you are faced with (assuming that you have mastered Roux style) a 3 move algorithm which will orient all 6 edges right then and there, vs. doing a setup for my system, it makes way more sense to use it than to trudge along with mine. You already know what to do, and the rest of the solve is trivial.

When you orient using Roux, what is the worst case number of Look-(U, U', or U2)-Execute iterations you can run into?

What is the average number of iterations?

I see that the best case scenario is one iteration.

Yeah, that's how it works, at first it seems stupid and wasteful. But once you use it a while you'll naturally have all of the U adjustments memorized and you'll realize that the moves are exactly same as memorized sequences but much more free. Having them separated like that allows you to control where UL/UR goes while you are orienting. Why have M'UM'U2M'UM' memorized when you can do a variation of M'UM' U2 M'U'M so that you can place UL/UR while orienting.

I am not exactly sure how you can control where UL and UR go while going through the process, as there are only 3 sequences to do (M'U2M', M2, M'UM'). I can see that you should be able to track where UL and or UR are going, but not control them. Are you talking about the optimizations listed at the bottom of Roux's page?

Roux shows nine patterns. I have eleven, which includes reflections. Do you do reflections, or do you adjust the cube to transform?

In the end, I whipped up my method to reduce the number of algorithms needed to do a reasonable speedcubing finish for corners first solves, a perspective which is making it harder for me than necessary to absorb the Roux concept.

In generic terms, it takes 21 algorithms to finish the L and R layers + orient the m edges with a typical corners first solve. This is where one layer is solved, and the last edge needs to be placed in the opposite layer:

LU edge is at DF: 8 cases, plus their mirrors = 16 algorithms
LU is in place but flipped: 2 cases = 2 algorithms
LU is solved when RU is solved: 2 adjacent M edges flipped, 2 diagonal, all 4 = 3 algorithms.

This doesn't include special cases where 2 or 4 M edges are flipped in place.

I like to think that I potentially improved things a bit by knocking off 10 algorithms.

If there are any useful bits in my system which might complement Roux, I will be pleased.

 

[Athefre]

I notice you have a new post, I started working on this one immediately after I posted my last (Two hours ago!). So I'll edit certain parts.

Average. Bold is your solutions, bold and italicized are the variations or shortcuts.

Scramble 1: M' U' M' U M U M U2 M' U2 M2 U' M U2 M2 U M U M' U' M' U' M U2 M
Yours: M' U M' U2 M U' M U2 M U' M U' M' U M' U' M' U2 (18)
Roux: U M' U2 M' U2 M U' M' U' M' U2 M U' M U2 M U2 (17)

Scramble 2: M U2 M U' M' U' M U2 M U2 M U2 M U2 M' U2 M' U' M' U' M U' M' U' M
Yours: U M M U' M' U M U2 M' U M U2 M (13)
Roux: M' U' M U M' U M' U2 M U' F2 M' F2 M (14)

Scramble 3: M' U' M2 U' M' U' M' U' M2 U2 M2 U' M U' M2 U' M2 U2 M U' M2 U M' U M2
Yours: U M2 U2 M U' M U2 M U' M U' M' U M U2 M2 U D2 M' D2 M (21)
Roux: M' U M' U2 M U' M U' M' U2 M' U' M U2 M U2 (16)

Scramble 4: M2 U2 M U M U2 M2 U' M2 U M' U' M U M' U M U' M' U' M' U' M2 U' M
Yours: U' M' M' U' M' U M U' M U' M2 U F2 M F2 M (16)
Roux: M2 U' M' U M' U M2 U D2 M D2 M (12)

Scramble 5: M2 U2 M U M2 U M2 U' M' U2 M' U M' U2 M U M U M U M U2 M' U M2
Yours: U2 M U M U' M' U M U2 M' U2 M2 U' F2 M' F2 M (17)
Roux: U M' U' M U M2 U M U2 M U2 (11)

Scramble 6: M2 U2 M' U M U2 M' U2 M2 U' M U2 M' U2 M' U' M2 U' M U' M U2 M' U M2
Yours: U M U' M U' M' U M U2 M' U' M2 U M2 (14)
Roux: U M' U M' U M' U2 M' U M U2 M U2 (13)

Scramble 7: M2 U2 M2 U' M' U M' U' M U2 M U2 M U2 M U' M U2 M2 U2 M U' M U M
Yours: M U' M M U M U M U M U' M2 U' M' U2 M (16)
Roux: M' U' M' U' M U' M' U D2 M' D2 M' (12)

Scramble 8: M2 U' M U' M U2 M U' M U2 M' U' M U M' U' M2 U2 M U' M2 U2 M' U' M'
Yours: U' M' U M' U M U' M' U2 M U M2 U' M' (14)
Roux: U2 M' U' M' U M' U2 M' U2 (9)

Scramble 9: M' U2 M2 U M' U' M2 U2 M' U M U' M2 U M U2 M2 U2 M' U M2 U2 M U M2
Yours: U' M' M' U M U2 M' U M U' M2 U F2 M2 F2 M2 (16)
Roux: M' U' M' U2 M' U' M U2 M2 U E2 M E2 M' (14)

Scramble 10: M U2 M2 U2 M2 U2 M2 U' M U' M2 U' M2 U2 M2 U2 M' U' M U M2 U' M2 U M'
Yours: U' M' U' M U' M' U M U2 M U M' U2 M' U2 (15)
Roux: M' U M' U M U' M' U' M2 U' M U2 M (13)

Scramble 11: M' U' M U2 M U2 M U M' U M U M' U M2 U M' U2 M' U2 M2 U M' U' M
Yours: MM' U M' U2 M' U M' U M U' M' U' M2 U E2 M E2 M' (19)
Roux: U' M' U' M' U2 M U' M' U' M2 U M' U2 M' U' (15)

Scramble 12: M2 U M' U M' U' M U2 M2 U' M' U' M' U2 M2 U M U' M' U M U2 M U2 M'
Yours: U M M U' M U M' U2 M U' M U' M U2 M2 U' M' U2 M2 (19)
Roux: M U M' U' M U' M' U2 M' U' F2 M' F2 M2 (14)

Your system: 18 (13) (21) 16 17 14 16 14 16 13 19 19
Average: 16.2

"Roux": (17) 14 16 12 11 13 12 (9) 14 13 15 14
Average: 13.4

Notes:

1. The "Roux" results were just whatever I saw first; just like a speedsolve, and if a scramble had misoriented centers and ended in an M/M' I removed it, if It ended in U/U'/U2/M2 I added a M' so that the centers would be oriented. This isn't cheating because my (and probably all "Roux" users) centers are always oriented at the start of the LSE because I do an r/r' as I place the final pair in the second block or I do an r/r'/l/l' during an R/R'/L/L' in a corner sequence if possible.

2. I know ones like number 6 could have been shortened at move number 3 but it's hard to notice something like that when going full speed.

3. Scramle 3's solution is way too long, U' M2 M' U M' U2 M' U M' U M U' M' U D2 M' D2 M would have been shorter but how would someone have known?

4. Are you doing variations of your solutions when possible? like M U' M U' M U' M instead of U'....M U M U M U M. This will shorten your average by a move or two.

Roux shows nine patterns. I have eleven, which includes reflections. Do you do reflections, or do you adjust the cube to transform?

Is note number 4 what you are asking? I had a hard time wording mine (didn't think to say "reflections")

I think that if faced with a situation where you are faced with (assuming that you have mastered Roux style) a 3 move algorithm which will orient all 6 edges right then and there, vs. doing a setup for my system, it makes way more sense to use it than to trudge along with mine. You already know what to do, and the rest of the solve is trivial.

Did you mean "orient all 4 edges"? That example I posted has 4 misoriented edges (UL/UB/UR/UF). I wish there was a 3 move sequence to orient 6 edges. The shortest I've ever found is 5...things like M'U'MUM', but you can't always use that to solve 6 edges, it depends on where UL/UR are.

When you orient using Roux, what is the worst case number of Look-(U, U', or U2)-Execute iterations you can run into?

What is the average number of iterations?

I see that the best case scenario is one iteration.

Almost all have two. The worst (if you don't use the unconstrained center optimizations) is when all 6 are misoriented, it requires three (it's the only one with three). If you use unconstrained centers then it's only two. It's important to know that not all of those optimizations on his page are useful when solving for speed. I use the 6 most easily recognizable and am considering adding more.

I am not exactly sure how you can control where UL and UR go while going through the process, as there are only 3 sequences to do (M'U2M', M2, M'UM'). I can see that you should be able to track where UL and or UR are going, but not control them.

Examples:

1. Scramble: M2 U' M U2 M2 U2 M U M' U2 M U' M2 U2 M2 U' M2 U' M U2 M2 U' M' U M2
Solution: On his page find the image with two edges next to each other misoriented and two misoriented at DF and DB. It says to do M2. Now you have 3 misoriented at UF/UR/UB and one misoriented at DB, the image for that exact pattern doesn't exist but there is a similar one that says to do M'UM' so just change that slightly and do U' MU'M. Notice that instead of MU'M there you could have done MUM or MU'M'. Any of those would have oriented the edges. In a speedsolve I would have chosen MU'M because it places a UL/UR edge in it's correct position.

2. M U M' U2 M U2 M2 U M2 U M U M' U' M U M U2 M2 U M' U2 M' U M'

Solution: For this pattern you are supposed to do M'U2M' first. Then the pattern that comes up says to do M'UM'. But notice that when you do that it puts a UL/UR edge at DF, instead what you should have done is M'UM. See how it puts that edge in it's correct position? Not having a set list of sequences allows you much more freedom to be able to control where you want what edge to go.

In the end, I whipped up my method to reduce the number of algorithms needed to do a reasonable speedcubing finish for corners first solves, a perspective which is making it harder for me than necessary to absorb the Roux concept.

In generic terms, it takes 21 algorithms to finish the L and R layers + orient the m edges with a typical corners first solve. This is where one layer is solved, and the last edge needs to be placed in the opposite layer:

LU edge is at DF: 8 cases, plus their mirrors = 16 algorithms
LU is in place but flipped: 2 cases = 2 algorithms
LU is solved when RU is solved: 2 adjacent M edges flipped, 2 diagonal, all 4 = 3 algorithms.

This doesn't include special cases where 2 or 4 M edges are flipped in place.

I like to think that I potentially improved things a bit by knocking off 10 algorithms.

If there are any useful bits in my system which might complement Roux, I will be pleased.

It seems to be a good alternative, I'm going to study it to see what parts are most easily recognizable for me to use in my speedsolves along with the Gilles' LSE.

 

[cubacca1972]

Looking at the results of your test, it seems that my method adds quite a few moves at the beginning (and during some of the longer algorithms) in order to achieve a "brute force" solution.

I will need to look closer at the Roux system, get over my aversion to the look-adjust-move part of it in order to get a better understanding of it.

When I said orient all 6 edges with 3 moves, I was referring to the whole task, and not the particular case with all 6 edges flipped (wouldn't that be nice!).

I think I also found a few 5 move algorithms which flipped lots of edges, but as you pointed out, they are useful only when UL and UR are in specific positions. This is probably why I started with the idea of parking UL and UR in the D layer.

If I understand you correctly, you are using variants of the M'U2M' and M'UM' algorithms during the orientation phase, as well as orienting your centers (U or D center of a solved cube is positioned in the U face). I will try to do the same.

I will try to get all 6 edges oriented in a maximum of 2 iterations, except for the all 6 flipped case.

I will post here if I run into problems.

 

[Zeroknight]

I have no idea what roux method is, but looking at this:

After this, you'll get an EPLL case 11/36 (30.6%) of the time, or a 4-move M-slice edge permutation 11/18 (61.1%) of the time, or a skip 1/12 (8.3%) of the time.

it suddenly becomes very attractive...

But wait are thes the only cases?

 

[Athefre]

Looking at the results of your test, it seems that my method adds quite a few moves at the beginning (and during some of the longer algorithms) in order to achieve a "brute force" solution.

I believe there are some shortcuts to be found. If you keep practicing you could probably shorten the average by 1-2 moves.

I will need to look closer at the Roux system, get over my aversion to the look-adjust-move part of it in order to get a better understanding of it.

When I said orient all 6 edges with 3 moves, I was referring to the whole task, and not the particular case with all 6 edges flipped (wouldn't that be nice!).

I think I also found a few 5 move algorithms which flipped lots of edges, but as you pointed out, they are useful only when UL and UR are in specific positions. This is probably why I started with the idea of parking UL and UR in the D layer.

That's neat because I found a few that were shorter:

M U' M U M' U2 M U' M U' M (11,12) - UF/UB
Alternative:

M U M' U2 M U M' (7,8) - UF/UR
Alternative:

M' U2 M' U2 M' U' M U2 M U2 M (11,15) - UL/UF/UR/UB
Alternative:

M' U M' U2 M' U M' U M U' M' (11,12) - UF/DF
Alternative: MU2M'UMUMU2M (9)

M' U M U' M' U2 M (7,8) - UL/UF/UR/DF
Alternative:

M U' M U2 M U' M U' M' U M (11,12) - UB/DB
Alternative: M'U2MU'M'U'M'U2M' (9)

M U' M' U M U2 M' (7,8) - UL/UB/UR/DB
Alternative: M'UMUM' (5)

M' U' M' U M U' M (7,7) - DF/DB
Alternative:

M' U2 M' U2 M U' M' U' M2 (9,12) - UL/UR/DF/DB
Alternative: M'U2M'U2M' (5) - If you were using Gilles system, it's only one (!) move to orient edges here - M'

M U M U M U M (7,7) - UF/UR/DF/DB
Alternative:

M' U M' U M' U M2 U' M U2 M' U' M (13,15) - UL/UF/UR/UB/DF/DB
Alternative:

I didn't bother much with the 7 move solutions, I focused on the ones with 9 moves or more. I wish I could have found something better for the 13 move case

I'll keep thinking about it.

If I understand you correctly, you are using variants of the M'U2M' and M'UM' algorithms during the orientation phase, as well as orienting your centers (U or D center of a solved cube is positioned in the U face). I will try to do the same.

I will try to get all 6 edges oriented in a maximum of 2 iterations, except for the all 6 flipped case.

I will post here if I run into problems.

 

[cubacca1972]

I believe there are some shortcuts to be found. If you keep practicing you could probably shorten the average by 1-2 moves.



That's neat because I found a few that were shorter:

M U' M U M' U2 M U' M U' M (11,12) - UF/UB
Alternative:

M U M' U2 M U M' (7,8) - UF/UR
Alternative:

M' U2 M' U2 M' U' M U2 M U2 M (11,15) - UL/UF/UR/UB
Alternative:

M' U M' U2 M' U M' U M U' M' (11,12) - UF/DF
Alternative: MU2M'UMUMU2M (9)

M' U M U' M' U2 M (7,8) - UL/UF/UR/DF
Alternative:

M U' M U2 M U' M U' M' U M (11,12) - UB/DB
Alternative: M'U2MU'M'U'M'U2M' (9)

M U' M' U M U2 M' (7,8) - UL/UB/UR/DB
Alternative: M'UMUM' (5)

M' U' M' U M U' M (7,7) - DF/DB
Alternative:

M' U2 M' U2 M U' M' U' M2 (9,12) - UL/UR/DF/DB
Alternative: M'U2M'U2M' (5) - If you were using Gilles system, it's only one (!) move to orient edges here - M'

M U M U M U M (7,7) - UF/UR/DF/DB
Alternative:

M' U M' U M' U M2 U' M U2 M' U' M (13,15) - UL/UF/UR/UB/DF/DB
Alternative:

I didn't bother much with the 7 move solutions, I focused on the ones with 9 moves or more. I wish I could have found something better for the 13 move case

I'll keep thinking about it.

I am speechless. When I did your 5 move alternative, my eyes nearly popped out! This is because it shaves 2 moves off an already short algorithm, but also because that is one of the 11 cases which has a mirror (my case 5).

The drop from 9 moves to 5 moves in case 9 is also astounding.

So far, this drops the average algorithm length from 9.18 moves to 8.09 moves!

So I have to ask: How did you find these? I used Ron's sticker mode solver to generate the original algorithms. I wonder if I have been using it to its fullest (Use the check centers function, or not? Wrong starting position for the 3 U layer edges flipped, 1 D layer edge flipped). Are you figuring these out manually?

 

[EmersonHerrmann]

Hmmm...you have many ideas Mr. Blah